CBSE Class 8 Maths Exemplar Chapter 4: Linear Equations in One Variable NCERT Solutions

Let's solve each equation to determine the nature of its solution:

1. For the equation $3x\; +\; 2\; =\; 5x\; +\; 2$:

   Subtract 2 from both sides: $3x\; =\; 5x$

   Subtract $3x$ from both sides: $0\; =\; 2x$

   Divide by 2: $x\; =\; 0$. This is an integer.

2. For the equation $4x\; -\; 18\; =\; 2$:

   Add 18 to both sides: $4x\; =\; 2\; +\; 18$

   $4x\; =\; 20$

   Divide by 4: $x\; =\; \backslash frac\{20\}\{4\}$

   $x\; =\; 5$. This is an integer.

3. For the equation $4x\; +\; 7\; =\; x\; +\; 2$:

   Subtract $x$ from both sides: $4x\; -\; x\; +\; 7\; =\; 2$

   $3x\; +\; 7\; =\; 2$

   Subtract 7 from both sides: $3x\; =\; 2\; -\; 7$

   $3x\; =\; -5$

   Divide by 3: $x\; =\; \backslash frac\{-5\}\{3\}$. This is a rational number, which is neither a positive nor negative integer, and is often described as 'neither a fraction nor an integer' in contexts distinguishing whole numbers from other rational numbers.

4. For the equation $5x\; -\; 8\; =\; x\; +\; 4$:

   Subtract $x$ from both sides: $5x\; -\; x\; -\; 8\; =\; 4$

   $4x\; -\; 8\; =\; 4$

   Add 8 to both sides: $4x\; =\; 4\; +\; 8$

   $4x\; =\; 12$

   Divide by 4: $x\; =\; \backslash frac\{12\}\{4\}$

   $x\; =\; 3$. This is an integer.

Therefore, the solution to the equation $4x\; +\; 7\; =\; x\; +\; 2$ is $\backslash frac\{-5\}\{3\}$, which fits the description.

We are given the linear equation $ax\; +\; b\; =\; 0$, where $a\; \backslash neq\; 0$.

To find the solution for $x$, we first isolate the term containing $x$. We transpose the constant term $b$ to the right-hand side (RHS) of the equation. When $b$ is moved from the left side to the right side, its sign changes from positive to negative.

$$ax\; =\; -b$$

Now, to solve for $x$, we divide both sides of the equation by the coefficient of $x$, which is $a$.

$$x\; =\; \backslash frac\{-b\}\{a\}$$

Thus, the solution to the equation $ax\; +\; b\; =\; 0$ is $x\; =\; \backslash frac\{-b\}\{a\}$.

We are given the equation $8x\; -\; 3\; =\; 25\; +\; 17x$.

To solve for $x$, we will use the transposition method. We move the terms involving $x$ to one side and the constant terms to the other side.

Let's move the constant term $-3$ from the left-hand side (LHS) to the right-hand side (RHS). When $-3$ is transposed, it becomes $+3$.

Let's move the variable term $17x$ from the RHS to the LHS. When $17x$ is transposed, it becomes $-17x$.

The equation becomes:

$$8x\; -\; 17x\; =\; 25\; +\; 3$$

Combine the like terms on both sides:

$$-9x\; =\; 28$$

Now, to find the value of $x$, divide both sides by $-9$:

$$x\; =\; \backslash frac\{28\}\{-9\}$$

$$x\; =\; -\backslash frac\{28\}\{9\}$$

The number $-\backslash frac\{28\}\{9\}$ is a rational number because it can be expressed as a ratio of two integers ($-28$ and $9$), where the denominator is not zero. It is not an integer (a whole number) and while it is a fraction, the term 'rational number' is the most precise classification.

Therefore, $x$ is a rational number.

In algebra, when we solve an equation, we often need to move terms from one side of the equals sign to the other. This process involves changing the sign of the term being moved. For example, if a term is added on one side, it is subtracted on the other side, and vice versa. Similarly, if a term is multiplied on one side, it is divided on the other side, and vice versa.

This operation of moving a term from one side of an equation to the other while changing its sign is specifically called **Transposition**.

Distributivity, Commutativity, and Associativity are properties related to the operations of addition and multiplication, not the movement of terms in an equation.

Therefore, the correct answer is Transposition.

We are given the equation $\backslash frac\{5x\}\{3\}\; -\; 4\; =\; \backslash frac\{2x\}\{5\}$. Our goal is to find the value of the expression $2x\; -\; 7$.

First, let's solve the given equation for $x$. We need to isolate the terms with $x$ on one side and the constant terms on the other.

Subtract $\backslash frac\{2x\}\{5\}$ from both sides and add $4$ to both sides:

$$\backslash frac\{5x\}\{3\}\; -\; \backslash frac\{2x\}\{5\}\; =\; 4$$

To combine the fractions on the left side, we find a common denominator, which is $15$ (the least common multiple of $3$ and $5$).

$$\backslash frac\{5x\; \backslash times\; 5\}\{3\; \backslash times\; 5\}\; -\; \backslash frac\{2x\; \backslash times\; 3\}\{5\; \backslash times\; 3\}\; =\; 4$$

$$\backslash frac\{25x\}\{15\}\; -\; \backslash frac\{6x\}\{15\}\; =\; 4$$

Combine the numerators:

$$\backslash frac\{25x\; -\; 6x\}\{15\}\; =\; 4$$

$$\backslash frac\{19x\}\{15\}\; =\; 4$$

Now, multiply both sides by $15$ to isolate $19x$:

$$19x\; =\; 4\; \backslash times\; 15$$

$$19x\; =\; 60$$

Finally, divide both sides by $19$ to find the value of $x$:

$$x\; =\; \backslash frac\{60\}\{19\}$$

Now that we have the value of $x$, we can find the value of the expression $2x\; -\; 7$.

Substitute $x\; =\; \backslash frac\{60\}\{19\}$ into the expression:

$$2x\; -\; 7\; =\; 2\backslash left(\backslash frac\{60\}\{19\}\backslash right)\; -\; 7$$

$$=\; \backslash frac\{120\}\{19\}\; -\; 7$$

To subtract $7$, we express it as a fraction with a denominator of $19$:

$$=\; \backslash frac\{120\}\{19\}\; -\; \backslash frac\{7\; \backslash times\; 19\}\{19\}$$

$$=\; \backslash frac\{120\}\{19\}\; -\; \backslash frac\{133\}\{19\}$$

$$=\; \backslash frac\{120\; -\; 133\}\{19\}$$

$$=\; \backslash frac\{-13\}\{19\}$$

Thus, the numerical value of $2x\; -\; 7$ is $-\backslash frac\{13\}\{19\}$.